# ring

• Jul 27th 2009, 01:11 PM
peteryellow
ring
Show that an ideal is proper if and only if it does not contain 1.
• Jul 27th 2009, 03:01 PM
siclar
Prove these two statements: If 1 is in the ideal, it is not proper. If 1 is not in the ideal, it is proper. The second statement is obvious by the definition of being a proper ideal, i.e. if the ring is $R$ and ideal is $I$, $I\neq R$ since $1\in R, 1 \notin I$.

Use the definition of an (left) ideal: for all elements $r$ of the ring, $rI\subseteq I$. In particular, if $i\in I$ then $ri\in I$ for any $r$. What if $i=1$?
• Jul 27th 2009, 09:07 PM
Gamma
Does your ring definition require all rings to be unitary?

If you take the ring to be $2\mathbb{Z}$. Certainly $2\mathbb{Z}\subset 2\mathbb{Z}$ and it does not contain 1, but is not a proper ideal since it is the whole ring.